1. **Problem statement:** Customer calls arrive at a call center following a Poisson process with a mean rate of 10 calls per hour. Let $T$ be the interarrival time between two consecutive calls. 2. **Part (a): PDF and CDF of $T$** - The interarrival time $T$ in a Poisson process is exponentially distributed with rate $\lambda = 10$ calls/hour. - The PDF (probability density function) of $T$ is: $$f_T(t) = \lambda e^{-\lambda t} = 10 e^{-10 t}, \quad t \geq 0$$ - The CDF (cumulative distribution function) of $T$ is: $$F_T(t) = P(T \leq t) = 1 - e^{-\lambda t} = 1 - e^{-10 t}, \quad t \geq 0$$ 3. **Part (b): Probability next call arrives in less than 3 minutes** - Convert 3 minutes to hours: $3 \text{ minutes} = \frac{3}{60} = 0.05$ hours. - Use the CDF: $$P(T < 0.05) = F_T(0.05) = 1 - e^{-10 \times 0.05} = 1 - e^{-0.5}$$ - Calculate: $$1 - e^{-0.5} \approx 1 - 0.6065 = 0.3935$$ - So, the probability is approximately 0.3935. 4. **Part (c): Median interarrival time** - The median $m$ satisfies: $$P(T \leq m) = 0.5$$ - Using the CDF: $$1 - e^{-10 m} = 0.5 \implies e^{-10 m} = 0.5$$ - Taking natural logarithm: $$-10 m = \ln(0.5) = -\ln(2)$$ $$m = \frac{\ln(2)}{10} \approx \frac{0.6931}{10} = 0.06931 \text{ hours}$$ - Convert to minutes: $$0.06931 \times 60 \approx 4.16 \text{ minutes}$$ 5. **Part (d): Probability wait more than 12 minutes given a call just came in** - Memoryless property of exponential distribution means: $$P(T > t) = e^{-\lambda t}$$ - Convert 12 minutes to hours: $$12 \text{ minutes} = \frac{12}{60} = 0.2 \text{ hours}$$ - Calculate: $$P(T > 0.2) = e^{-10 \times 0.2} = e^{-2} \approx 0.1353$$ **Final answers:** - (a) PDF: $f_T(t) = 10 e^{-10 t}$, CDF: $F_T(t) = 1 - e^{-10 t}$ - (b) $P(T < 3 \text{ min}) \approx 0.3935$ - (c) Median interarrival time $\approx 4.16$ minutes - (d) $P(T > 12 \text{ min}) \approx 0.1353$